Solution Set Chapter 4 & 5
Problems: 4-32, 4-34, 4-36, 4-41; 5-28, 5-30, 5-32
4.32
a)
Convert demand to annual basis:  = (380)(12) = 4560 per year,
c = 0.45, h = Ic = (.25)(.45) = .1125, K = 8.50.
Q* =
b)
2K

h
(2)(8.5)(4560)
.1125
= 830
T = Q*/ = 830/4560
= .1820 years. (2.184 months)
  T
r = (2/12)4560 = 760
760

c)
Profit per sale = .99 - .45 = .54
Total annual profit = (.54)(4560) = $2462.40
Subtract annual holding and set-up cost:
2Kh 
(2)(8.50)(4560)(.1125)
= 93.39
Hence the annual profit = $2369.01
4.34
K

c
a)
=
=
=
=
200 (both cases)
(200)(52) = 10,400
75 locally
65 overseas
Average annual cost at the optimal solution is G(Q) = c +
2KIc in both cases. Substituting the appropriate costs we obtain:
locally
G(Q) = (10,400)(75) +
(2)(200)(10,400)(.2)(75)
= $787,899.37
overseas
G(Q) = (10,400)(65) +
(2)(200)(10,400)(.2)(65)
= $683,353.91
Difference = $104,545 annually.
b)
The average pipeline inventory is  . Hence:
locally:
(200)(52)(1/12)
=
866.67 units
overseas:
(200)(52)(6/12)
=
5,200 units
The value of the local inventory is (866.67)(75) = $65,000
Value of overseas is (5,200)(65) = $338,000
Difference = $273,000
Assuming 20% rate of interest, this difference would cost the firm
(273,000)(.2) = $54,600 annually, which is less than the savings realized
in holding and set-up cost realized by producing overseas. Hence on the
basis of cost alone, overseas production is still preferred.
c)
The differences in cost of $50,000 yearly could easily be outweighed by
the firm's competitive advantage at being able to be more responsive to
market demands due to the shortened production lead time locally.
4.36 The President used the following EOQ:
Q1 =
= 707
(2)(100)(1800)
(.3)(2.40)
"True" EOQ was:
Q2
=
(2)(40)(1800)
(,.2)(2.40)
= 548
Cost error in percent =
1  Q Q*  1 707 548  =



2 
Q * Q 
 2 548 707 
1.033
Error is 3.3% only
Optimal cost =
= 263
(2)(40)(1800)(.2)(2.40)
Additional cost = 263(.033) = $8.68 annually
4.41
a)
c = .35 + .15 + .05 = .55
h = Ic = .11
EOQ
b)
2K

h
Q
50,000

Q* 120,604
(2)(400)(2,000,000)
(.11)
= 120,604
= .41458
Error = 0.5[Q/Q* + Q*/Q] = 1.4133  41.33% additional cost
Optimal cost
=
2Kh 
=
(2)(400)(2, 000, 000)(.11)
13,266.5
~
$13,267 annually.
Hence, the additional cost of using suboptimal policy:
= (13,267)(.4133) = $5483 annually.
5.28
a) Newsboy model. Each three-month period corresponds to a period. Since
there is considerable variation from one three-month period to the next, we
must assume random demand.
b) Since the demand rate is constant, an EOQ model would be appropriate.
c) Because of the monthly variation, a random demand model is appropriate.
The perishability does not appear to be an issue so that a (Q,R) model is
probably appropriate. Whether to use a stock-out or service level model is
unclear.
d) EOQ with quantity discounts.
e) Other type of model. This problem would require a model which includes
both quantity discounts and random demand. Such a model was not
considered in this chapter.
5.30
a) F(R) = .95  z = 1.645
R = z + µ = (102.1)(1.645) + 166.67 = 335
b) n(R)/Q = 1 -  = .05
n(R) = (.05)(EOQ) = (.05)(1741) = 87.05
L(z) = n(R)/ = 87.05/102.1 = .8526
z = -.71
R = z +  = (102.1)(-.71) + 166.67 = 94
c) n(R0) = L(z) = (102.1)(.8526) = 87.05
1 - F(R0)
=
.7612
2
Q1 = 8705  (1741) 2  87.05
.7612
L(z1) =
.7612
= 1859
(.05)(1859)
= .910
102.1
z1 = -.79
R1 = (102.1)(-.79) + 166.67 = 86
n(R1) = L(z) = (102.1)(.910) = 92.91
1 - F(R1) = .7853
Q2
92.91
92.912
 (1741)2 
.7853
.7853
=
L(z2) =
z2
=
(.05)(1863)
102.1
=
=
1863
.912
-.79 same as z1.
Stop.
(Q,R) = (1863,86)
5.32
a)
Newsboy problem
cu = 50 - 20 = 30
c0 = 20 - 8 + .3(20) = 18
Critical ratio = 30/(18 + 30) = .625
Demand is uniform from 20 to 70
.625
20
Q  20
70  20
b)
 =
Q
= .625  Q = 51
70  20
2
= 45
 = 7
70
Q* = z +  where F(z) = .625.
From the table we obtain z = .32.
Hence, Q* = (7)*(.32) + 45  47